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Автор - Riko787

Решите дифференциальное уравнение

2xy′ + y=3x

Заранее огромное спасибо

Ответ

Проверено экспертом

Автор - NNNLLL54

2xy'+y=3x\\y'=frac{3x-y}{2x}\\t=frac{y}{x}; ,; ; y=tx; ,; ; y'=t'x+t\\t'x+t=frac{3x-tx}{2x}\\t'x+t=frac{3}{2}-frac{t}{2}\\t'x=frac{3}{2}-frac{3, t}{2}\\t'=frac{dt}{dx} =frac{3-3t}{2x}\\int frac{dt}{3-3t}=int frac{dx}{2x}\\-frac{1}{3}, ln|3-3t|=frac{1}{2}, ln|x|+lnC\\lnsqrt[3]{3-3t}=-ln(Csqrt{x})\\sqrt[3]{3-3cdot frac{y}{x}}=frac{1}{Csqrt{x}}\\3cdot (1-frac{y}{x})=frac{1}{C^3sqrt{x^3}}\\frac{y}{x}=1-frac{1}{3C^3sqrt{x^3}}\\y=xcdot (1-frac{1}{3C^3sqrt{x^3}})

Ответ

Автор - BanahSt

Ответ:

y = x + C/√x

Пошаговое объяснение:

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